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Consumer Mathematics: Calculating and Comparing Simple Interest and Compound Interest

Mary deposits [tex]$\$[/tex]30,000[tex]$ into an account that pays $[/tex]4\%[tex]$ interest per year, compounded annually. Josh deposits $[/tex]\[tex]$30,000$[/tex] into an account that also pays [tex]$4\%$[/tex] per year, but it is simple interest.

Find the interest Mary and Josh earn during each of the first three years. Then decide who earns more interest for each year. Assume there are no withdrawals and no additional deposits.

\begin{tabular}{|c|c|c|}
\hline Year & \begin{tabular}{l}
Interest Mary earns \\
(Interest compounded annually)
\end{tabular} & \begin{tabular}{l}
Interest Josh earns \\
(Simple interest)
\end{tabular} \\
\hline First & [tex]$\$[/tex] \square[tex]$ & $[/tex]\[tex]$ \square$[/tex] \\
\hline Second & [tex]$\$[/tex] \square[tex]$ & $[/tex]\[tex]$ \square$[/tex] \\
\hline Third & [tex]$\$[/tex] \square[tex]$ & $[/tex]\[tex]$ \square$[/tex] \\
\hline
\end{tabular}


Sagot :

To determine the interest earned by Mary and Josh during each of the first three years, we need to use the appropriate formulas for compound interest and simple interest respectively.

### Given Data:
- Principal [tex]\( P = \$30,000 \)[/tex]
- Interest Rate [tex]\( r = 4\% = 0.04 \)[/tex]

### Compound Interest (Mary)
The formula for compound interest when compounded annually is:
[tex]\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Since the interest is compounded annually, [tex]\( n = 1 \)[/tex], the formula simplifies to:
[tex]\[ A = P \times (1 + r)^t \][/tex]

### Simple Interest (Josh)
The formula for simple interest is:
[tex]\[ I = P \times r \times t \][/tex]

Now, let's calculate the interest for each of the first three years.

#### Year 1

Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^1 = 30,000 \times 1.04 = 31,200 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 31,200 - 30,000 = 1,200 \][/tex]

Josh:
[tex]\[ I = 30,000 \times 0.04 \times 1 = 1,200 \][/tex]

#### Year 2

Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^2 = 30,000 \times 1.0816 = 32,448 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 32,448 - 30,000 = 2,448 \][/tex]

Josh:
[tex]\[ I = 30,000 \times 0.04 \times 2 = 2,400 \][/tex]

#### Year 3

Mary:
[tex]\[ A = 30,000 \times (1 + 0.04)^3 = 30,000 \times 1.124864 = 33,745.92 \][/tex]
Interest earned:
[tex]\[ \text{Interest} = 33,745.92 - 30,000 = 3,745.92 \][/tex]

Josh:
[tex]\[ I = 30,000 \times 0.04 \times 3 = 3,600 \][/tex]

### Summary:

The interest earned by Mary and Josh during each of the first three years is summarized in the table below:

[tex]\[\begin{array}{|c|c|c|} \hline \text{Year} & \text{Interest Mary earns} & \text{Interest Josh earns} \\ \hline \text{First} & \$1,200 & \$1,200 \\ \hline \text{Second} & \$2,448 & \$2,400 \\ \hline \text{Third} & \$3,745.92 & \$3,600 \\ \hline \end{array}\][/tex]

### Conclusion:

Who earns more interest?
- In the first year, both Mary and Josh earn the same amount of interest: [tex]$1,200. - In the second and third years, Mary earns more interest than Josh. Specifically, in the second year she earns $[/tex]2,448 compared to Josh's [tex]$2,400, and in the third year, she earns $[/tex]3,745.92 compared to Josh's $3,600.